Should be: parmenides51 wrote: How many of the numbers $1\cdot 2\cdot 3,2\cdot 3\cdot 4,\ldots,2020 \cdot 2021 \cdot 2022$ are divisible by $2020$?

Disclaimer: The problem is bad and I'm never doing a Sweden MO problem again. We proceed with bashing. Because $2020 = 20 \cdot 101$ we only need to bash out $20$ terms (the multiples of $101$). \begin{tabular}{ |c|c| } \hline Multiple of 101 & Number of triplets satisfying the condition containing the number \\ [0.5ex] \hline 101 & 2 \\ 202 & 1 \\ 303 & 1 \\ 404 & 2 \\ 505 & 2 \\ 606 & 1 \\ 707 & 0 \\ 808 & 1 \\ 909 & 1 \\ 1010 & 2 \\ 1111 & 1 \\ 1212 & 1 \\ 1313 & 0 \\ 1414 & 1 \\ 1515 & 2 \\ 1616 & 2 \\ 1717 & 1 \\ 1818 & 1 \\ 1919 & 2 \\ 2020 & 3 \\ \hline \end{tabular}Summing these numbers up we get our answer of $27$.

ilikemath40 wrote: Disclaimer: The problem is bad The problem isn't bad, you just did it in a bad way.

jasperE3 wrote: ilikemath40 wrote: Disclaimer: The problem is bad The problem isn't bad, you just did it in a bad way. I asked other people and they said that the best way was just to bash. If you have a better solution, post it!

All we have to do is find the number of $n$ such that $2020|n(n+1)(n+2)$. Note that $2020=2^{2}\cdot 5\cdot 101$ $2020|n(n+1)(n+2)\Leftrightarrow2^{2}|n(n+1)(n+2), 5|n(n+1)(n+2),101|n(n+1)(n+2)\Leftrightarrow n\equiv 0,2,3(\text{mod 4}),n\equiv 0,3,4(\text{mod 5}),n\equiv 0,99,100(\text{mod 101})$ By C.R.T, we get the answer 27